If \( \sin A=\frac{2 \sqrt{6}}{5} \) and \( A \in\left[90^{\circ} ; 360^{\circ}\right] \) calculate without the use of a calculator and with the aid of a diagram the value of \( 15 \tan A \cdot \cos A \) If \( \sin \alpha=\frac{3}{5} \) with \( \alpha \in\left[00^{\circ}\right. \)

We are given that   sin A = (2√6)/5               (1) with A in the interval [90°, 360°]. We are asked to find the value of   15 · tan A · cos A without using a calculator and by means of a diagram. Below is a step‐by‐step solution. ──────────────────────── Step 1. Find cos A from sin A. A diagram of a right–triangle (or using the Pythagorean identity) is most helpful. Recall that   sin²A + cos²A = 1. We have   sin²A = [(2√6)/5]² = (4·6)/25 = 24/25. Thus,   cos²A = 1 – 24/25 = 1/25, so that   |cos A| = 1/5. However, the sign of cos A depends on the quadrant in which A lies. Since sin A > 0 but the interval [90°, 360°] includes the 2nd and 3rd quadrants, we must decide which quadrant A is in. Notice that sin A = (2√6)/5 is positive; in standard position the sine is positive in the 1st and 2nd quadrants. Since A is between 90° and 360°, the only acceptable quadrant is the 2nd quadrant. In the 2nd quadrant, cosine is negative. Therefore, we have   cos A = –1/5. ──────────────────────── Step 2. Find tan A from sin A and cos A. Recall that   tan A = sin A / cos A. Thus,   tan A = [(2√6)/5] / (–1/5) = –2√6. ──────────────────────── Step 3. Compute 15 · tan A · cos A. Substitute the values found:   tan A = –2√6  and  cos A = –1/5. So,   15 · tan A · cos A = 15 · (–2√6) · (–1/5). Multiply the negatives: (–2√6) · (–1/5) = 2√6/5. Then,   15 · (2√6/5) = (15/5) · 2√6 = 3 · 2√6 = 6√6. ──────────────────────── Thus, the value of 15 · tan A · cos A is 6√6. ──────────────────────── Part 2: Additional Information The problem statement also gives:   sin α = 3/5  with  α ∈ [0°, …) (At this point the interval seems to be intended as for an acute angle, and often such problems continue asking for related quantities.) For an acute angle α (that is, 0° ≤ α < 90°), we can draw a right–angled triangle whose angle α satisfies   sin α = (Opposite) / (Hypotenuse) = 3/5. Using the Pythagorean theorem, the side adjacent to α is   Adjacent = √(Hypotenuse² – Opposite²) = √(5² – 3²) = √(25 – 9) = √16 = 4. Thus, for α:   cos α = Adjacent / Hypotenuse = 4/5,   tan α = Opposite / Adjacent = 3/4. ──────────────────────── Summary of Answers 1. Given sin A = (2√6)/5 with A in [90°, 360°]:   cos A = –1/5,   tan A = –2√6,   15 · tan A · cos A = 6√6. 2. Given sin α = 3/5 with α acute:   cos α = 4/5,   tan α = 3/4. These results have been obtained without using a calculator and with the aid of diagrams as appropriate.